What is the solution to this system of linear equations? 12q + 3r = 15 –4q – 4r = –44 (–18, 29) (–2, 13) (8, –1) (15, –44)

12q +3r =15    
3(4q +r) = 3*5  
Divide both sides by 3
4q +r = 5
Time for the second one:
-4q -4r = -44  
-4(q +r) = (-4)*11  
Divide both sides by -4
 q +r = 11
4q +r = 5
 q +r = 11
Subtract the second from first equation 
3q 0 = -6
3q = -6
q = -6/3 = -2
q = -2
q +r = 11
-2 +r =11
r = 11 +2 
r = 13 
q = -2
r = 13
Your answer is...
(-2,13)